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261
CHAPTER 9
Section 9.1 1.
a. ( ) ( ) ( ) 4.5.41.4 −=−=−=− YEXEYXE , irrespective of sample sizes.
b. ( ) ( ) ( ) ( ) ( )0724.
1000.2
1008.1 222
22
1 =+=+=+=−nm
YVXVYXVσσ
, and the s.d.
of 2691.0724. ==− YX . c. A normal curve with mean and s.d. as given in a and b (because m = n = 100, the CLT
implies that both X and Y have approximately normal distributions, so YX − does also). The shape is not necessarily that of a normal curve when m = n = 10, because the CLT cannot be invoked. So if the two lifetime population distributions are not normal,
the distribution of YX − will typically be quite complicated.
2. The test statistic value is
ns
ms
yxz
22
21 +
−= , and Ho will be rejected if either 96.1≥z or
96.1−≤z . We compute 85.433.433
2100
45
1900
452200
400,40500,4222
==
+
−=z . Since 4.85 >
1.96, reject Ho and conclude that the two brands differ with respect to true average tread lives.
3. The test statistic value is ( )
ns
ms
yxz
22
21
5000
+
−−= , and Ho will be rejected at level .01 if
33.2≥z . We compute ( )
76.193.396
700
45
1500
452200
5000800,36500,4322
==
+
−−=z , which is not
> 2.33, so we don’t reject Ho and conclude that the true average life for radials does not exceed that for economy brand by more than 500.
Chapter 9: Inferences Based on Two Samples
262
4. a. From Exercise 2, the C.I. is
( ) ( ) ( ) 33.849210033.43396.1210096.122
21 ±=±=+±−
ns
ms
yx
( )33.2949,67.1250= . In the context of this problem situation, the interval is moderately wide (a consequence of the standard deviations being large), so the information about 1µ and 2µ is not as precise as might be desirable.
b. From Exercise 3, the upper bound is
( ) 95.635295.652570093.396645.15700 =+=+ . 5.
a. Ha says that the average calorie output for sufferers is more than 1 cal/cm2/min below that
for nonsufferers. ( ) ( )
1414.1016.
1004. 222
221 =+=+
nmσσ
, so
( ) ( )90.2
1414.105.264.
−=−−−
=z . At level .01, Ho is rejected if 33.2−≤z ; since –
2.90 < 2.33, reject Ho.
b. ( ) 0019.90.2 =−Φ=P
c. ( ) 8212.92.11414.
12.133.21 =−Φ−=
+−
−−Φ−=β
d. ( )
( )15.65
2.28.133.22.
2
2
=−
+== nm , so use 66.
Chapter 9: Inferences Based on Two Samples
263
6.
a. Ho should be rejected if 33.2≥z . Since ( )
33.253.3
3296.1
4056.2
87.1612.18≥=
+
−=z , Ho
should be rejected at level .01.
b. ( ) ( ) 3085.50.3539.
0133.21 =−Φ=
−
−Φ=β
c. ( )
06.370529.96.1
1169.28.1645.1
196.14056.2
2=⇒=⇒=
+=+ n
nn, so use
n = 38. d. Since n = 32 is not a large sample, it would no longer be appropriate to use the large
sample test. A small sample t procedure should be used (section 9.2), and the appropriate conclusion would follow.
7.
1 Parameter of interest: =− 21 µµ the true difference of means for males and
females on the Boredom Proneness Rating. Let =1µ men’s average and =2µ women’s average.
2 Ho: 021 =− µµ
3 Ha: 021 >− µµ
4 ( ) ( )
ns
ms
yx
ns
ms
yxz o
22
21
22
21
0
+
−−=
+
∆−−=
5 RR: 645.1≥z
6 ( )
83.1
14868.4
9783.4
26.940.1022
=
+
∆−−= oz
7 Reject Ho. The data indicates the Boredom Proneness Rating is higher for males than for females.
Chapter 9: Inferences Based on Two Samples
264
8. a.
1 Parameter of interest: =− 21 µµ the true difference of mean tensile strength of the
1064 grade and the 1078 grade wire rod. Let =1µ 1064 grade average and =2µ 1078 grade average.
2 Ho: 1021 −=− µµ
3 Ha: 1021 −<− µµ
4 ( ) ( ) ( )
ns
ms
yx
ns
ms
yxz o
22
21
22
21
10
+
−−−=
+
∆−−=
5 RR: α<− valuep
6 ( ) ( )
57.28210.
6
1290.2
1293.1
106.1236.10722
−=−
=
+
−−−=z
7 For a lowertailed test, the pvalue = ( ) 057.28 ≈−Φ , which is less than any α , so reject Ho. There is very compelling evidence that the mean tensile strength of the 1078 grade exceeds that of the 1064 grade by more than 10.
b. The requested information can be provided by a 95% confidence interval for 21 µµ − :
( ) ( ) ( ) ( )588.5,412.6210.96.1696.122
21 −−=±−=+±−
ns
ms
yx .
9.
a. point estimate 2.67.139.19 =−=− yx . It appears that there could be a difference. b.
Ho: 021 =− µµ ,Ha: 021 ≠− µµ , ( )
14.144.52.6
608.15
601.39
7.139.1922
==
+
−=z , and
the pvalue = 2[P(z > 1.14)] = 2( .1271) = .2542. The p value is larger than any reasonable α, so we do not reject H0. There is no significant difference.
c. No. With a normal distribution, we would expect most of the data to be within 2 standard deviations of the mean, and the distribution should be symmetric. 2 sd’s above the mean is 98.1, but the distribution stops at zero on the left. The distribution is positively skewed.
d. We will calculate a 95% confidence interval for µ, the true average length of stays for
patients given the treatment. ( )8.21,0.109.99.1960
1.3996.19.19 =±=±
Chapter 9: Inferences Based on Two Samples
265
10. a. The hypotheses are Ho: 521 =− µµ and Ha: 521 >− µµ . At level .001, Ho should
be rejected if 08.3≥z . Since ( )
08.389.22272.
58.596.65<=
−−=z , Ho cannot be
rejected in favor of Ha at this level, so the use of the high purity steel cannot be justified.
b. 121 =∆−− oµµ , so ( ) 2891.53.2272.
108.3 =−Φ=
−Φ=β
11. ( )ns
ms
zYX22
21
2/ +±− α . Standard error = ns
. Substitution yields
( ) ( ) ( )22
212/ SESEzyx +±− α . Using ,05.=α 96.12/ =αz , so
( ) ( ) ( ) ( )41.2,99.02.03.096.18.35.5 22 =+±− . Because we selected ,05.=α we
can state that when using this method with repeated sampling, the interval calculated will bracket the true difference 95% of the time. The interval is fairly narrow, indicating precision of the estimate.
12. The C.I. is ( ) ( ) 46.277.89104.58.277.858.222
21 ±−=±−=+±−
ns
ms
yx
( )31.6,23.11 −−= . With 99% confidence we may say that the true difference between the average 7day and 28day strengths is between 11.23 and 6.31 N/mm2.
13. 05.21 == σσ , d = .04, 05.,01. == βα , and the test is onetailed, so
( )( )38.49
0016.645.133.20025.0025. 2
=++
=n , so use n = 50.
14. The appropriate hypotheses are Ho: 0=θ vs. Ha: 0<θ , where 212 µµθ −= . ( 0<θ is
equivalent to 212 µµ < , so normal is more than twice schizo) The estimator of θ is
YX −= 2θ̂ , with ( ) ( ) ( )nm
YVarXVarVar22
214
4ˆ σσθ +=+= , θσ is the square root
of ( )θ̂Var , and θσ̂ is obtained by replacing each 2iσ with 2
iS . The test statistic is then
θσθ
ˆ
ˆ (since 0=oθ ), and Ho is rejected if .33.2−≤z With ( ) 97.35.669.22ˆ −=−=θ
and ( ) ( )
9236.4503.4
433.24ˆ
22
=+=θσ , 05.19236.
97.−=
−=z ; Because –1.05 > 2.33,
Ho is not rejected.
Chapter 9: Inferences Based on Two Samples
266
15.
a. As either m or n increases, σ decreases, so σµµ o∆−− 21 increases (the numerator is
positive), so
∆−−
−σµµ
αoz 21 decreases, so
∆−−
−Φ=σµµ
β αoz 21
decreases.
b. As β decreases, βz increases, and since βz is the numerator of n , n increases also.
16.
nns
ns
yxz
22.
22
21
=
+
−= . For n = 100, z = 1.41 and pvalue = ( )[ ] 1586.41.112 =Φ− .
For n = 400, z = 2.83 and pvalue = .0046. From a practical point of view, the closeness of x and y suggests that there is essentially no difference between true average fracture toughness for type I and type I steels. The very small difference in sample averages has been magnified by the large sample sizes – statistical rather than practical significance. The pvalue by itself would not have conveyed this message.
Section 9.2 17.
a. ( )
( ) ( )1743.17
44.1694.21.37
99
2
1062
105
2
106
105
22
22
≈=+
=
+
+=ν
b. ( )
( ) ( )217.21
411.694.01.24
149
2
1562
105
2
156
105
22
22
≈=+
=
+
+=ν
c. ( )
( ) ( )1827.18
411.018.84.7
149
2
1562
102
2
156
102
22
22
≈=+
=
+
+=ν
d. ( )
( ) ( )2605.26
098.395.84.12
2311
2
2462
125
2
246
125
22
22
≈=+
=
+
+=ν
Chapter 9: Inferences Based on Two Samples
267
18. With Ho: 021 =− µµ vs. Ha: 021 ≠− µµ , we will reject Ho if α<− valuep .
( )( ) ( )
68.6
45
2
5240.2
6164.
2
5240.
6164.
22
22
≈=
+
+=ν , and the test statistic
17.61265.78.95.2173.22
5240.
6164. 22
==+
−=t leads to a pvalue of 2[ P(t > 6.17)] < 2(.0005) =.001,
which is less than most reasonable s'α , so we reject Ho and conclude that there is a difference in the densities of the two brick types.
19. For the given hypotheses, the test statistic 20.1007.3
6.3103.1297.115
638.5
603.5 22
−=−
=+
+−=t , and
the d.f. is ( )
( ) ( )96.9
58241.4
52168.4
8241.42168.422
2
=+
+=ν , so use d.f. = 9. We will reject Ho if
;764.29,01. −=−≤ tt since –1.20 > 2.764, we don’t reject Ho.
20. We want a 95% confidence interval for 21 µµ − . 262.29,025. =t , so the interval is
( ) ( )20.3,40.10007.3262.26.3 −=±− . Because the interval is so wide, it does not appear that precise information is available.
21. Let =1µ the true average gap detection threshold for normal subjects, and =2µ the
corresponding value for CTS subjects. The relevant hypotheses are Ho: 021 =− µµ vs.
Ha: 021 <− µµ , and the test statistic 46.23329.
82.07569.0351125.
53.271.1−=
−=
+−
=t .
Using d.f. ( )
( ) ( )1.15
907569.
70351125.
07569.0351125.22
2
=+
+=ν , or 15, the rejection region is
602.215,01. −=−≤ tt . Since –2.46 is not 602.2−≤ , we fail to reject Ho. We have
insufficient evidence to claim that the true average gap detection threshold for CTS subjects exceeds that for normal subjects.
Chapter 9: Inferences Based on Two Samples
268
22. Let =1µ the true average strength for wirebrushing preparation and let =2µ the average strength for handchisel preparation. Since we are concerned about any possible difference between the two means, a twosided test is appropriate. We test 0: 210 =− µµH vs.
0: 21 ≠− µµaH . We need the degrees of freedom to find the rejection region:
( )( ) ( )
33.141632.0039.
3964.2
1111
2
501.42
1258.1
2
1201.4
1258.1
22
22
=+
=
+
+=ν , which we round down to 14, so we
reject Ho if 145.214,025. =≥ tt . The test statistic is
( ) 159.32442.1
93.313.2320.19
1201.4
1258.1 22
−=−
=+
−=t , which is 145.2−≤ , so we reject Ho and
conclude that there does appear to be a difference between the two population average strengths.
23.
a. Normal plots
Using Minitab to generate normal probability plots, we see that both plots illustrate sufficient linearity. Therefore, it is plausible that both samples have been selected from normal population distributions.
PValue: 0.344ASquared: 0.396
AndersonDarling Normality Tes t
N: 24StDev : 0.444206Average: 1.50833
2.31.81.30.8
.999
.99
.95
.80
.50
.20
.05
.01
.001
Pro
bab
ility
H:
Normal Probability Plot for High Quali ty Fabric
Average: 1. 58750St Dev : 0.530330N: 24
AndersonDarling Normality Tes tASquared: 10.670PValue: 1. 000
1.0 1.5 2.0 2.5
.001
.01
.05
.20
.50
.80
.95
.99
.999
Pro
babi
lity
P :
Normal Probability Plot for Poor Quality Fabric
Chapter 9: Inferences Based on Two Samples
269
b.
0.5 1.5 2.5
Comparative Box Plot for High Quality and Poor Quality Fabric
QualityPoor
QualityHigh
extensibility (%)
The comparative boxplot does not suggest a difference between average extensibility for the two types of fabrics.
c. We test 0: 210 =− µµH vs. 0: 21 ≠− µµaH . With degrees of freedom
( )5.10
00017906.0433265. 2
==ν , which we round down to 10, and using significance level
.05 (not specified in the problem), we reject Ho if 228.210,025. =≥ tt . The test
statistic is ( )
38.0433265.
08.−=
−=t , which is not 228.2≥ in absolute value, so we
cannot reject Ho. There is insufficient evidence to claim that the true average extensibility differs for the two types of fabrics.
24. A 95% confidence interval for the difference between the true firmness of zeroday apples
and the true firmness of 20day apples is ( )2039.
2066.
96.474.822
,025. +±− νt . We
calculate the degrees of freedom ( ) ( )
83.30
1919
2039.
2066.
2
2039.2
2066.
222
22=
+
+
=ν , so we use 30 df, and
042.230,025. =t , so the interval is ( ) ( )13.4,43.317142.042.278.3 =± . Thus, with
95% confidence, we can say that the true average firmness for zeroday apples exceeds that of 20day apples by between 3.43 and 4.13 N.
Chapter 9: Inferences Based on Two Samples
270
25. We calculate the degrees of freedom ( )
( ) ( )95.53
3027
2
318.72
285.5
2
318.7
285.5
22
22
=
+
+=ν , or about 54 (normally
we would round down to 53, but this number is very close to 54 – of course for this large number of df, using either 53 or 54 won’t make much difference in the critical t value) so the
desired confidence interval is ( ) 318.7
285.5 22
68.13.885.91 +±−
( )131.6,269.931.22.3 =±= . Because 0 does not lie inside this interval, we can be
reasonably certain that the true difference 21 µµ − is not 0 and, therefore, that the two population means are not equal. For a 95% interval, the t value increases to about 2.01 or so, which results in the interval 506.32.3 ± . Since this interval does contain 0, we can no longer conclude that the means are different if we use a 95% confidence interval.
26. Let =1µ the true average potential drop for alloy connections and let =2µ the true average potential drop for EC connections. Since we are interested in whether the potential drop is higher for alloy connections, an upper tailed test is appropriate. We test 0: 210 =− µµH
vs. 0: 21 >− µµaH . Using the SAS output provided, the test statistic, when assuming
unequal variances, is t = 3.6362, the corresponding df is 37.5, and the pvalue for our upper
tailed test would be ½ (twotailed pvalue) = ( ) 0004.0008.21 = . Our pvalue of .0004 is
less than the significance level of .01, so we reject Ho. We have sufficient evidence to claim that the true average potential drop for alloy connections is higher than that for EC connections.
27. The approximate degrees of freedom for this estimate are
( )( ) ( )
83.8175.10159.893
75
2
83.82
63.11
2
83.8
63.11
22
22
==
+
+=ν , which we round down to 8, so 306.28,025. =t
and the desired interval is ( ) ( )4674.5306.29.18306.24.213.40 83.8
63.11 22
±=+±−
( )5.31,3.6607.129.18 =±= . Because 0 is not contained in this interval, there is strong
evidence that 21 µµ − is not 0; i.e., we can conclude that the population means are not equal.
Calculating a confidence interval for 12 µµ − would change only the order of subtraction of the sample means, but the standard error calculation would give the same result as before. Therefore, the 95% interval estimate of 12 µµ − would be ( 31.5, 6.3), just the negatives of the endpoints of the original interval. Since 0 is not in this interval, we reach exactly the same conclusion as before; the population means are not equal.
Chapter 9: Inferences Based on Two Samples
271
28. We will test the hypotheses: 10: 210 =− µµH vs. 10: 21 >− µµaH . The test
statistic is ( )( )
08.217.25.410
544.4
1075.2 22
==+
−−=
yxt The degrees of freedom
( )( ) ( )
659.595.308.22
49
2
544.42
1075.2
2
544.4
1075.2
22
22
≈==
+
+=ν and the pvalue from table A.8 is approx .04,
which is < .10 so we reject H0 and conclude that the true average lean angle for older females is more than 10 degrees smaller than that of younger females.
29. Let =1µ the true average compression strength for strawberry drink and let =2µ the true average compression strength for cola. A lower tailed test is appropriate. We test
0: 210 =− µµH vs. 0: 21 <− µµaH . The test statistic is
10.2154.29
14−=
+−
=t . ( )
( ) ( )3.25
8114.7736.1971
1415
144.29
4.4422
2
==+
=ν , so use df=25.
The pvalue 023.)10.2( =−<≈ tP . This pvalue indicates strong support for the
alternative hypothesis. The data does suggest that the extra carbonation of cola results in a higher average compression strength.
30.
a. We desire a 99% confidence interval. First we calculate the degrees of freedom:
( )( ) ( )
24.37
2626
2
263.42
262.2
2
263.4
262.2
22
22
=
+
+=ν , which we would round down to 37, except that there is
no df = 37 row in Table A.5. Using 36 degrees of freedom (a more conservative choice),
719.236,005. =t , and the 99% C.I. is
( ) ( )83.6,98.11576.24.9719.28.424.33 263.4
262.2 22
−−=±−=+±− . We are
very confident that the true average load for carbon beams exceeds that for fiberglass beams by between 6.83 and 11.98 kN.
b. The upper limit of the interval in part a does not give a 99% upper confidence bound. The 99% upper bound would be ( ) 09.79473.434.24.9 −=+− , meaning that the true average load for carbon beams exceeds that for fiberglass beams by at least 7.09 kN.
Chapter 9: Inferences Based on Two Samples
272
31. a.
The mo st notable feature of these boxplots is the larger amount of variation present in the midrange data compared to the highrange data. Otherwise, both look reasonably symmetric with no outliers present.
b. Using df = 23, a 95% confidence interval for rangehighrangemid −− − µµ is
( ) ( )54.9,84.769.885.069.245.4373.438 1183.6
171.15 22
−=±=+±− . Since
plausible values for rangehighrangemid −− − µµ are both positive and negative (i.e., the
interval spans zero) we would conclude that there is not sufficient evidence to suggest that the average value for midrange and the average value for highrange differ.
32. Let =1µ the true average proportional stress limit for red oak and let =2µ the true average
proportional stress limit for Douglas fir. We test 1: 210 =− µµH vs. 1: 21 >− µµaH .
The test statistic is ( )
818.12084.83.1165.648.8
1028.1
1479. 22
=+
−−=t . With degrees of freedom
( )( ) ( )
1485.13
913
2084.2
1028.12
1479.
2
22≈=
+
=ν , the pvalue 046.)8.1( =>≈ tP . This pvalue
indicates strong support for the alternative hypothesis since we would reject Ho at significance levels greater than .046. There is sufficient evidence to claim that true average proportional stress limit for red oak exceeds that of Douglas fir by more than 1 MPa.
high rangem id range
470
460
450
440
430
420m
id r
ange
Comparative Box Plot for High Range and Mid Range
Chapter 9: Inferences Based on Two Samples
273
33. Let =1µ the true average weight gain for steroid treatment and let =2µ the true average weight gain for the population not treated with steroids. The exercise asks if we can conclude that 2µ exceeds 1µ by more than 5 g., which we can restate in the equivalent form:
521 −<− µµ . Therefore, we conduct a lowertailed test of 5: 210 −=− µµH vs.
5: 21 −<− µµaH . The test statistic is
( ) ( ) ( )2.223.2
2124.17.2
105.2
86.2
55.408.32222
221
≈−=−
=
+
−−−=
+
∆−−=
ns
ms
yxt . The approximate d.f. is
( )( ) ( )
876.141454.1609.2
97
2
105.22
86.2
2
105.2
86.2
22
22
==
+
+=ν , which we round down to 14. The pvalue for a
lower tailed test is P( t < 2.2 ) = P( t > 2.2 ) = .022. Since this pvalue is larger than the specified significance level .01, we cannot reject Ho. Therefore, this data does not support the belief that average weight gain for the control group exceeds that of the steroid group by more than 5 g.
34.
a. Following the usual format for most confidence intervals: statistic ± (critical value)(standard error), a pooled variance confidence interval for the difference between
two means is ( ) nmpnm styx 112,2/ +⋅±− −+α .
b. The sample means and standard deviations of the two samples are 90.13=x ,
225.11 =s , 20.12=y , 010.12 =s . The pooled variance estimate is =2ps
( ) ( )2222
21 010.1
24414
225.1244
142
12
1
−+−
+
−+−
=
−+−
+
−+−
snm
ns
nmm
260.1= , so 1227.1=ps . With df = m+n1 = 6 for this interval, 447.26,025. =t and
the desired interval is ( ) ( )( ) 41
411227.1447.220.1290.13 +±−
( )64.3,24.943.17.1 −=±= . This interval contains 0, so it does not support the conclusion that the two population means are different.
c. Using the twosample t interval discussed earlier, we use the CI as follows: First, we need
to calculate the degrees of freedom. ( )
( ) ( )919.9
0686.6302.
33
2
401.12
4225.1
2
401.1
4225.1
22
22
≈==
+
+=ν so
262.29,025. =t . Then the interval is
( ) ( ) ( )50.3,10.7938.262.270.1262.22.129.13 401.1
4225.1 22
−=±=+±− . This
interval is slightly smaller, but it still supports the same conclusion.
Chapter 9: Inferences Based on Two Samples
274
35. There are two changes that must be made to the procedure we currently use. First, the
equation used to compute the value of the t test statistic is: ( ) ( )
nms
yxt
p11
+
∆−−= where sp is
defined as in Exercise 34 above. Second, the degrees of freedom = m + n – 2. Assuming equal variances in the situation from Exercise 33, we calculate sp as follows:
( ) ( ) 544.25.2169
6.2167 22 =
+
=ps . The value of the test statistic is, then,
( ) ( )2.224.2
101
81
544.2
55.408.32−≈−=
+
−−−=t . The degrees of freedom = 16, and the p
value is P ( t < 2.2) = .021. Since .021 > .01, we fail to reject Ho. This is the same conclusion reached in Exercise 33.
Section 9.3 36. 25.7=d , 8628.11=Ds
1 Parameter of Interest: =Dµ true average difference of breaking load for fabric in unabraded or abraded condition.
2 0:0 =DH µ
3 0: >DaH µ
4 ns
dns
dt
DD
D
/0
/−
=−
=µ
5 RR: 998.27,01. =≥ tt
6 73.18/8628.11
025.7=
−=t
7 Fail to reject Ho. The data does not indicate a difference in breaking load for the two fabric load conditions.
Chapter 9: Inferences Based on Two Samples
275
37. a. This exercise calls for paired analysis. First, compute the difference between indoor and
outdoor concentrations of hexavalent chromium for each of the 33 houses. These 33
differences are summarized as follows: n = 33, 4239.−=d , 3868.=ds , where d =
(indoor value – outdoor value). Then 037.232,025. =t , and a 95% confidence interval
for the population mean difference between indoor and outdoor concentration is
( ) ( )2868.,5611.13715.4239.33
3868.037.24239. −−=±−=
±− . We can be
highly confident, at the 95% confidence level, that the true average concentration of hexavalent chromium outdoors exceeds the true average concentration indoors by between .2868 and .5611 nanograms/m3.
b. A 95% prediction interval for the difference in concentration for the 34th house is
( ) ( )( ) ( )3758,.224.113868.037.24239.1 3311
32,025. −=+±−=+± ndstd .
This prediction interval means that the indoor concentration may exceed the outdoor concentration by as much as .3758 nanograms/m3 and that the outdoor concentration may exceed the indoor concentration by a much as 1.224 nanograms/m3, for the 34th house. Clearly, this is a wide prediction interval, largely because of the amount of variation in the differences.
38.
a. The median of the “Normal” data is 46.80 and the upper and lower quartiles are 45.55 and 49.55, which yields an IQR of 49.55 – 45.55 = 4.00. The median of the “High” data is 90.1 and the upper and lower quartiles are 88.55 and 90.95, which yields an IQR of 90.95 – 88.55 = 2.40. The most significant feature of these boxplots is the fact that their locations (medians) are far apart.
Normal :High:
90
80
70
60
50
40
Comparative Boxplots
for Normal and High Strength Concrete Mix
Chapter 9: Inferences Based on Two Samples
276
b. This data is paired because the two measurements are taken for each of 15 test conditions. Therefore, we have to work with the differences of the two samples. A quantile of the 15 differences shows that the data follows (approximately) a straight line, indicating that it is reasonable to assume that the differences follow a normal distribution. Taking
differences in the order “Normal” – “High” , we find 23.42−=d , and 34.4=ds .
With 145.214,025. =t , a 95% confidence interval for the difference between the
population means is
( ) ( )83.39,63.44404.223.421534.4
145.223.42 −−=±−=
±− . Because 0 is
not contained in this interval, we can conclude that the difference between the population means is not 0; i.e., we conclude that the two population means are not equal.
39.
a. A normal probability plot shows that the data could easily follow a normal distribution.
b. We test 0:0 =dH µ vs. 0: ≠daH µ , with test statistic
7.274.214/228
02.167/
0≈=
−=
−=
nsd
tD
. The twotailed pvalue is 2[ P( t > 2.7)] =
2[.009] = .018. Since .018 < .05, we can reject Ho . There is strong evidence to support the claim that the true average difference between intake values measured by the two methods is not 0. There is a difference between them.
40.
a. Ho will be rejected in favor of Ha if either 947.215,005. =≥ tt or 947.2−≤t . The
summary quantities are 544.−=d , and 714.=ds , so 05.31785.
544.−=
−=t .
Because 947.205.3 −≤− , Ho is rejected in favor of Ha.
b. 31.72 =ps , 70.2=ps , and 57.96.544.
−=−
=t , which is clearly insignificant; the
incorrect analysis yields an inappropriate conclusion.
41. We test 0:0 =dH µ vs. 0: >daH µ . With 600.7=d , and 178.4=ds ,
9.187.139.16.2
9/178.45600.7
≈==−
=t . With degrees of freedom n – 1 = 8, the
corresponding pvalue is P( t > 1.9 ) = .047. We would reject Ho at any alpha level greater than .047. So, at the typical significance level of .05, we would (barely) reject Ho, and conclude that the data indicates that the higher level of illumination yields a decrease of more than 5 seconds in true average task completion time.
Chapter 9: Inferences Based on Two Samples
277
42. 1 Parameter of interest: dµ denotes the true average difference of spatial ability in
brothers exposed to DES and brothers not exposed to DES. Let
.expexp osedunosedd µµµ −=
2 0:0 =DH µ
3 0: <DaH µ
4 ns
dns
dt
DD
D
/0
/−
=−
=µ
5 RR: Pvalue < .05, df = 8
6 ( )
2.25.0
07.136.12−=
−−=t , with corresponding pvalue .029 (from Table A.8)
7 Reject Ho. The data supports the idea that exposure to DES reduces spatial ability. 43.
a. Although there is a “jump” in the middle of the Normal Probability plot, the data follow a reasonably straight path, so there is no strong reason for doubting the normality of the population of differences.
b. A 95% lower confidence bound for the population mean difference is:
( ) 14.4954.1060.381518.23
761.160.3814,05. −=−−=
−−=
−
n
std d .
Therefore, with a confidence level of 95%, the population mean difference is above (–49.14).
c. A 95% upper confidence bound for the corresponding population mean difference is
14.4954.1060.38 =+ 44. We need to check the differences to see if the assumption of normality is plausible. A
probability chart will validate our use of the t distribution. A 95% confidence interval:
( ) 91.22263.263516645.508
753.163.263515,05. +=
+=
+
n
std d
( )54.2858,∞⇒ 45. The differences (white – black) are –7.62, 8.00, 9.09, 6.06, 1.39, 16.07, 8.40, 8.89, and
–2.88, from which 600.7−=d , and 178.4=ds . The confidence level is not specified in
the problem description; for 95% confidence, 306.28,025. =t , and the C.I. is
( ) ( )389.4,811.10211.3600.79
178.4306.2600.7 −−=±−=
±− .
46. With ( ) ( )5,6, 11 =yx , ( ) ( )14,15, 22 =yx , ( ) ( )0,1, 33 =yx , and ( ) ( )20,21, 44 =yx ,
1=d and 0=ds (the d I’s are 1, 1, 1, and 1), while s1 = s2 = 8.96, so sp = 8.96 and t = .16.
Chapter 9: Inferences Based on Two Samples
278
Section 9.4 47. Ho will be rejected if 33.201. −=−≤ zz . With 150.ˆ1 =p , and 300.ˆ 2 =p ,
263.800210
6002008030ˆ ==
++
=p , and 737.ˆ =q . The calculated test statistic is
( )( )( )18.4
0359.150.
737.263.
300.150.
6001
2001
−=−
=+
−=z . Because 33.218.4 −≤− , Ho is
rejected; the proportion of those who repeat after inducement appears lower than those who repeat after no inducement.
48.
a. Ho will be rejected if 96.1≥z . With 2100.30063ˆ1 ==p , and 4167.
18075ˆ 2 ==p ,
2875.1803007563ˆ =
++
=p , ( )( )( )
84.40427.2067.
7125.2875.
4167.2100.
1801
3001
−=−
=+
−=z .
Since 96.184.4 −≤− , Ho is rejected. b. 275.=p and 0432.=σ , so power =
( )( )[ ] ( )( )[ ]=
+−
Φ−
+
Φ−0432.
2.0421.96.10432.
2.0421.96.11
( ) ( )[ ] 9967.72.254.61 =Φ−Φ− .
49. 1 Parameter of interest: p1 – p2 = true difference in proportions of those responding to
two different survey covers. Let p1 = Plain, p2 = Picture. 2 0: 210 =− ppH
3 0: 21 <− ppH a
4 ( )nmqp
ppz
11
21
ˆˆ
ˆˆ
+
−=
5 Reject Ho if pvalue < .10
6 ( )( )( )
1910.213
12071
420207
420213
213109
207104
−=+
−=z ; pvalue = .4247
7 Fail to Reject Ho. The data does not indicate that plain cover surveys have a lower response rate.
Chapter 9: Inferences Based on Two Samples
279
50. Let 05.=α . A 95% confidence interval is ( ) ( )nqp
mqpzpp 2211 ˆˆˆˆ
2/21 ˆˆ +±− α
( ) ( )( ) ( )( ) ( )1708,.0160.0774.0934.266395
96.1 266140
266126
395171
395224
266126
395224 =±=
+±−= .
51.
a. 210 : ppH = will be rejected in favor of 21: ppH a ≠ if either 645.1≥z or
645.1−≤z . With 193.ˆ1 =p , and 182.ˆ 2 =p , 188.ˆ =p , 48.100742.
011.==z .
Since 1.48 is not 645.1≥ , Ho is not rejected and we conclude that no difference exists. b. Using formula (9.7) with p1 = .2, p2 = .18, 1.=α , 1.=β , and 645.12/ =αz ,
( )( )( )6582
0004.1476.16.28.162.138.5.645.1
2
=++
=n
52. Let p1 = true proportion of irradiated bulbs that are marketable; p2 = true proportion of
untreated bulbs that are marketable; The hypotheses are 0: 210 =− ppH vs.
0: 210 >− ppH . The test statistic is ( )nmqp
ppz
11
21
ˆˆ
ˆˆ
+
−= . With 850.
180153ˆ1 ==p , and
661.180119ˆ 2 ==p , 756.
360272ˆ ==p ,
( )( )( )2.4
045.189.
244.756.
661.850.
1801
1801
==+
−=z .
The pvalue = ( ) 02.41 ≈Φ− , so reject Ho at any reasonable level. Radiation appears to be beneficial.
53.
a. A 95% large sample confidence interval formula for ( )θln is
( )ny
ynmx
xmz
−+
−± 2/
ˆln αθ . Taking the antilogs of the upper and lower bounds
gives the confidence interval for θ itself.
b. 818.1ˆ037,11
104
034,11189
==θ , ( ) 598.ˆln =θ , and the standard deviation is
( )( ) ( )( ) 1213.104037,11
933,10189034,11
845,10=+ , so the CI for ( )θln is
( ) ( )836,.360.1213.96.1598. =± . Then taking the antilogs of the two bounds gives
the CI for θ to be ( )31.2,43.1 .
Chapter 9: Inferences Based on Two Samples
280
54. a. The “after” success probability is p1 + p3 while the “before” probability is p1 + p2 , so p1 +
p3 > p1 + p2 becomes p3 > p2; thus we wish to test 230 : ppH = versus
23: ppH a > .
b. The estimator of (p1 + p3) – (p1 + p2) is ( ) ( )
nXX
nXXXX 232131 −
=+−+
.
c. When Ho is true, p2 = p3, so n
ppn
XXVar 3223 +
=
−
, which is estimated by
npp 32 ˆˆ +
. The Z statistic is then 32
23
32
23
ˆˆ XXXX
npp
nXX
+−
=+
−
.
d. The computed value of Z is 68.2150200
150200=
+−
, so ( ) 0037.68.21 =Φ−=P . At
level .01, Ho can be rejected but at level .001 Ho would not be rejected.
55. 550.40
715ˆ1 =+
=p , 690.4229ˆ 2 ==p , and the 95% C.I. is
( ) ( ) ( )07,.35.21.14.106.96.1690.550. −=±−=±− .
56. Using p1 = q1 = p2 = q2 = .5, ( )nnn
L 7719.225.25.96.12 =
+= , so L=.1 requires n=769.
Section 9.5 57.
a. From Table A.9, column 5, row 8, 69.38,5,01. =F .
b. From column 8, row 5, 82.45,8,01. =F .
c. 207.1
5,8,05.8,5,95. ==
FF .
Chapter 9: Inferences Based on Two Samples
281
d. 271.1
8,5,05.5,8,95. ==
FF
e. 30.412,10,01. =F
f. 212.71.411
10,12,01.12,10,99. ===
FF .
g. 16.64,6,05. =F , so ( ) 95.16.6 =≤FP .
h. Since 177.64.51
5,10,99. ==F ,
( ) ( ) ( )177.74.474.4177. ≤−≤=≤≤ FPFPFP 94.01.95. =−= . 58.
a. Since the given f value of 4.75 falls between 33.310,5,05. =F and 64.510,5,01. =F , we
can say that the uppertailed pvalue is between .01 and .05. b. Since the given f of 2.00 is less than 52.210,5,10. =F , the pvalue > .10.
c. The two tailed pvalue = ( ) 02.)01(.264.52 ==≥FP . d. For a lower tailed test, we must first use formula 9.9 to find the critical values:
3030.1
5,10,10.10,5,90. ==
FF , 2110.
1
5,10,05.10,5,95. ==
FF ,
0995.1
5,10,01.10,5,99. ==
FF . Since .0995 < f = .200 < .2110, .01 < pvalue < .05 (but
obviously closer to .05). e. There is no column for numerator d.f. of 35 in Table A.9, however looking at both df =
30 and df = 40 columns, we see that for denominator df = 20, our f value is between F.01 and F.001. So we can say .001< pvalue < .01.
Chapter 9: Inferences Based on Two Samples
282
59. We test 22
0 21: σσ =H vs.
2221
: σσ ≠aH . The calculated test statistic is
( )( )
384.44.475.2
2
2
==f . With numerator d.f. = m – 1 = 10 – 1 = 9, and denominator d.f. = n –
1 = 5 – 1 = 4, we reject H0 if 00.64,9,05. =≥ Ff or
275.63.311
9,4,05.4,9,95. ===≤ FFf . Since .384 is in neither rejection region, we do
not reject H0 and conclude that there is no significant difference between the two standard deviations.
60. With =1σ true standard deviation for notfused specimens and =2σ true standard
deviation for fused specimens, we test 210 : σσ =H vs. 21: σσ >aH . The calculated
test statistic is ( )( )
814.19.2053.277
2
2
==f . With numerator d.f. = m – 1 = 10 – 1 = 9, and
denominator d.f. = n – 1 = 8 – 1 = 7, 7,9,10.72.2814.1 Ff =<= . We can say that the p
value > .10, which is obviously > .01, so we cannot reject Ho. There is not sufficient evidence that the standard deviation of the strength distribution for fused specimens is smaller than that of notfused specimens.
61. Let =21σ variance in weight gain for lowdose treatment, and =2
2σ variance in weight
gain for control condition. We wish to test 22
210 : σσ =H vs. 2
221: σσ >aH . The test
statistic is 22
21
ss
f = , and we reject Ho at level .05 if 08.222,19,05. ≈> Ff .
( )( )
8.2085.23254
2
2
≥==f , so reject Ho at level .05. The data does suggest that there is
more variability in the lowdose weight gains.
62. 210 : σσ =H will be rejected in favor of 21: σσ ≠aH if either 56.44,47,975. ≈≤ Ff
or if 8.144,47,025. ≈≥ Ff . Because 22.1=f , Ho is not rejected. The data does not
suggest a difference in the two variances.
Chapter 9: Inferences Based on Two Samples
283
63. ασσ
αα −=
≤≤ −−−−− 1
//
1,1,2/22
22
21
21
1,1,2/1 nmnm FSS
FP . The set of inequalities inside the
parentheses is clearly equivalent to 2
1
1,1,2/22
21
22
21
1,1,2/122
S
FS
S
FS nmnm −−−−− ≤≤ αα
σσ
. Substituting
the sample values 21s and 2
2s yields the confidence interval for 21
22
σσ
, and taking the square
root of each endpoint yields the confidence interval for 1
2
σσ
. m = n = 4, so we need
28.93,3,05. =F and 108.28.91
3,3,95. ==F . Then with s1 = .160 and s2 = .074, the C. I.
for 21
22
σσ
is (.023, 1.99), and for 1
2
σσ
is (.15, 1.41).
64. A 95% upper bound for 1
2
σσ
is ( ) ( )
( )10.8
79.
18.359.32
2
21
9,9,05.22 ==
s
Fs. We are
confident that the ratio of the standard deviation of triacetate porosity distribution to that of the cotton porosity distribution is at most 8.10.
Supplementary Exercises 65. We test 0: 210 =− µµH vs. 0: 21 ≠− µµaH . The test statistic is
( ) ( )22.3
524.1550
24150
1041
1027
757807222
221
===
+
−=
+
∆−−=
ns
ms
yxt . The approximate d.f. is
( )( ) ( )
6.15
91.168
99.72
24122
2
=+
=ν , which we round down to 15. The pvalue for a two
tailed test is approximately 2P( t > 3.22) = 2( .003) = .006. This small of a pvalue gives strong support for the alternative hypothesis. The data indicates a significant difference.
Chapter 9: Inferences Based on Two Samples
284
66. a.
Although the median of the fertilizer plot is higher than that of the control plots, the fertilizer plot data appears negatively skewed, while the opposite is true for the control plot data.
b. A test of 0: 210 =− µµH vs. 0: 21 ≠− µµaH yields a t value of .20, and a two
tailed pvalue of .85. (d.f. = 13). We would fail to reject Ho; the data does not indicate a significant difference in the means.
c. With 95% confidence we can say that the true average difference between the tree density
of the fertilizer plots and that of the control plots is somewhere between –144 and 120. Since this interval contains 0, 0 is a plausible value for the difference, which further supports the conclusion based on the pvalue.
67. Let p1 = true proportion of returned questionnaires that included no incentive; p2 = true
proportion of returned questionnaires that included an incentive. The hypotheses are
0: 210 =− ppH vs. 0: 210 <− ppH . The test statistic is ( )nmqp
ppz
11
21
ˆˆ
ˆˆ
+
−= .
682.11075ˆ1 ==p , and 673.
9866ˆ 2 ==p . At this point we notice that since 21 ˆˆ pp > , the
numerator of the z statistic will be > 0, and since we have a lower tailed test, the pvalue will be > .5. We fail to reject Ho. This data does not suggest that including an incentive increases the likelihood of a response.
Fe rtiliz Co ntrol
1000
1100
1200
1300
1400
Fer
tiliz
Comparative Boxplot of Tree Density BetweenFertilizer Plots and Control Plots
Chapter 9: Inferences Based on Two Samples
285
68. Summary quantities are m = 24, 66.103=x , s1 = 3.74, n = 11, 11.101=y , s2 = 3.60. We
use the pooled t interval based on 24 + 11 – 2 = 33 d.f.; 95% confidence requires
03.233,025. =t . With 68.132 =ps and 70.3=ps , the confidence interval is
( )( ) ( )28.5,18.73.255.270.303.255.2 111
241 −=±=+± . We are confident that the
difference between true average dry densities for the two sampling methods is between .18 and 5.28. Because the interval contains 0, we cannot say that there is a significant difference between them.
69. The center of any confidence interval for 21 µµ − is always 21 xx − , so
3.6092
9.16913.47321 =
+−=− xx . Furthermore, half of the width of this interval is
( )6.1082
23.4739.1691
=−−
. Equating this value to the expression on the right of the
95% confidence interval formula, ( )2
22
1
2196.16.1082
ns
ns
+= , we find
35.55296.1
6.1082
2
22
1
21 ==+
ns
ns
. For a 90% interval, the associated z value is 1.645, so
the 90% confidence interval is then ( )( ) 6.9083.60935.552645.13.609 ±=±
( )9.1517,3.299−= . 70.
a. A 95% lower confidence bound for the true average strength of joints with a side coating
is ( ) 78.5945.323.631096.5
833.123.639,025. =−=
−=
−
ns
tx . That is,
with a confidence level of 95%, the average strength of joints with a side coating is at least 59.78 (Note: this bound is valid only if the distribution of joint strength is normal.)
b. A 95% lower prediction bound for the strength of a single joint with a side coating is
( ) ( )( )1011
9,025. 196.5833.123.631 +−=+− nstx 77.5146.1123.63 =−= .
That is, with a confidence level of 95%, the strength of a single joint with a side coating would be at least 51.77.
c. For a confidence level of 95%, a twosided tolerance interval for capturing at least 95%
of the strength values of joints with side coating is ±x (tolerance critical value)s. The tolerance critical value is obtained from Table A.6 with 95% confidence, k = 95%, and n = 10. Thus, the interval is
( )( ) ( )37.83,09.4314.2023.6396.5379.323.63 =±=± . That is, we can be highly confident that at least 95% of all joints with side coatings have strength values between 43.09 and 83.37.
Chapter 9: Inferences Based on Two Samples
286
d. A 95% confidence interval for the difference between the true average strengths for the
two types of joints is ( ) ( ) ( )1096.5
1059.9
23.6395.8022
,025. +±− νt . The
approximate degrees of freedom is ( )
( ) ( )05.15
99
2105216.352
109681.91
2105216.35
109681.91
=
+
+=ν , so we use 15
d.f., and 131.215,025. =t . The interval is , then,
( )( ) ( )33.25,11.1061.772.1757.3131.272.17 =±=± . With 95% confidence, we can say that the true average strength for joints without side coating exceeds that of joints with side coating by between 10.11 and 25.33 lbin./in.
71. m = n = 40, 0.3975=x , s1 = 245.1, 0.2795=y , s2 = 293.7. The large sample 99%
confidence interval for 21 µµ − is ( )40
7.29340
1.24558.20.27950.3975
22
+±−
( ) ( )1336,10245.15600.1180 ≈± . The value 0 is not contained in this interval so we can
state that, with very high confidence, the value of 21 µµ − is not 0, which is equivalent to
concluding that the population means are not equal. 72. This exercise calls for a paired analysis. First compute the difference between the amount of
cone penetration for commutator and pinion bearings for each of the 17 motors. These 17
differences are summarized as follows: n = 17, 18.4−=d , 85.35=ds , where d =
(commutator value – pinion value). Then 120.216,025. =t , and the 95% confidence interval
for the population mean difference between penetration for the commutator armature bearing and penetration for the pinion bearing is:
( ) ( )25.14,61.2243.1818.41785.35
120.218.4 −=±−=
±− . We would have to say
that the population mean difference has not been precisely estimated. The bound on the error of estimation is quite large. In addition, the confidence interval spans zero. Because of this, we have insufficient evidence to claim that the population mean penetration differs for the two types of bearings.
Chapter 9: Inferences Based on Two Samples
287
73. Since we can assume that the distributions from which the samples were taken are normal, we use the twosample t test. Let 1µ denote the true mean headability rating for aluminum killed
steel specimens and 2µ denote the true mean headability rating for silicon killed steel. Then
the hypotheses are 0: 210 =− µµH vs. 0: 21 ≠− µµaH . The test statistic is
25.2086083.
66.047203.03888.
66.−=
−=
+−
=t . The approximate degrees of freedom
( )( ) ( )
5.57
29047203.
2903888.
086083.22
2
=+
=ν , so we use 57. The twotailed pvalue
( ) 028.014.2 =≈ , which is less than the specified significance level, so we would reject Ho. The data supports the article’s authors’ claim.
74. Let 1µ denote the true average tear length for Brand A and let 2µ denote the true average
tear length for Brand B. The relevant hypotheses are 0: 210 =− µµH vs.
0: 21 >− µµaH . Assuming both populations have normal distributions, the twosample t
test is appropriate. m = 16, 0.74=x , s1 = 14.8, n = 14, 0.61=y , s2 = 12.5, so the
approximate d.f. is ( )
( ) ( )97.27
1315
2
145.122
168.14
2
145.12
168.14
22
22
=
+
+=ν , which we round down to 27. The test
statistic is 6.20.610.74
145.12
168.14 22
≈+
−=t . From Table A.7, the pvalue = P( t > 2.6) = .007. At a
significance level of .05, Ho is rejected and we conclude that the average tear length for Brand A is larger than that of Brand B.
75.
a. The relevant hypotheses are 0: 210 =− µµH vs. 0: 21 ≠− µµaH . Assuming
both populations have normal distributions, the twosample t test is appropriate. m = 11, 1.98=x , s1 = 14.2, n = 15, 2.129=y , s2 = 39.1. The test statistic is
84.2252.1201.31
9207.1013309.181.31
−=−
=+
−=t . The approximate degrees of
freedom ( )
( ) ( )64.18
149207.101
103309.18
252.12022
2
=+
=ν , so we use 18. From Table A.7,
the twotailed pvalue ( ) 012.006.2 =≈ . No, obviously, the results are different.
Chapter 9: Inferences Based on Two Samples
288
b. For the hypotheses 25: 210 −=− µµH vs. 25: 21 −<− µµaH , the test statistic
changes to ( )
556.252.120
251.31−=
−−−=t . With degrees of freedom 18, the pvalue
( ) 278.6. =−<≈ tP . Since the pvalue is greater than any sensible choice of α , we
fail to reject Ho. There is insufficient evidence that the true average strength for males exceeds that for females by more than 25N.
76.
a. The relevant hypotheses are 0: 210 =− ∗∗ µµH (which is equivalent to saying
021 =− µµ ) versus 0: 21 ≠− ∗∗ µµaH (which is the same as saying
021 ≠− µµ ). The pooled t test is based on d.f. = m + n – 2 = 8 + 9 – 2 = 15. The
pooled variance is =2ps 2
221 2
12
1s
nmn
snm
m
−+−
+
−+−
( ) ( )22 6.4298
199.4
29818
−+−
+
−+−
49.22= , so 742.4=ps . The test statistic
is 0.304.3742.4
0.110.18**
91
8111
≈=+
−=
+
−=
nmps
yxt . From Table A.7, the pvalue
associated with t = 3.0 is 2P( t > 3.0 ) = 2(.004) = .008. At significance level .05, Ho is
rejected and we conclude that there is a difference between ∗1µ and ∗
2µ , which is
equivalent to saying that there is a difference between 1µ and 2µ .
b. No. The mean of a lognormal distribution is ( ) 2/2∗∗ += σµµ e , where
∗µ and ∗σ are
the parameters of the lognormal distribution (i.e., the mean and standard deviation of
ln(x)). So when ∗∗ = 21 σσ , then ∗∗ = 21 µµ would imply that 21 µµ = . However,
when ∗∗ ≠ 21 σσ , then even if ∗∗ = 21 µµ , the two means 1µ and 2µ (given by the
formula above) would not be equal. 77. This is paired data, so the paired t test is employed. The relevant hypotheses are
0:0 =dH µ vs. 0: <daH µ , where dµ denotes the difference between the population
average control strength minus the population average heated strength. The observed differences (control – heated) are: .06, .01, .02, 0, and .05. The sample mean and standard
deviation of the differences are 024.−=d and 0305.=ds . The test statistic is
8.176.1024.
50305.
−≈−=−
=t . From Table A.7, with d.f. = 5 – 1 = 4, the lower tailed p
value associated with t = 1.8 is P( t < 1.8) = P( t > 1.8 ) = .073. At significance level .05, Ho should not be rejected. Therefore, this data does not show that the heated average strength exceeds the average strength for the control population.
261
78. Let 1µ denote the true average ratio for young men and 2µ denote the true average ratio for
elderly men. Assuming both populations from which these samples were taken are normally distributed, the relevant hypotheses are 0: 210 =− µµH vs. 0: 21 >− µµaH . The
value of the test statistic is ( )( ) ( )
5.7
1228.
1322.
71.647.722
=
+
−=t . The d.f. = 20 and the pvalue is
P( t > 7.5) 0≈ . Since the pvalue is 05.=< α , we reject Ho. We have sufficient evidence to claim that the true average ratio for young men exceeds that for elderly men.
79.
101
4
3
2
1
Normal Score
Poo
r Vis
ibili
ty
101
2.5
1.5
0.5
Normal Score
Goo
d V
isib
ility
A normal probability plot indicates the data for good visibility does not follow a normal distribution, thus a ttest is not appropriate for this small a sample size.
80. The relevant hypotheses would be FM µµ = versus FM µµ ≠ for both the distress and delight indices. The reported pvalue for the test of mean differences on the distress index was less than 0.001. This indicates a statistically significant difference in the mean scores, with the mean score for women being higher. The reported pvalue for the test of mean differences on the delight index was > 0.05. This indicates a lack of statistical significance in the difference of delight index scores for men and women.
Chapter 9: Inferences Based on Two Samples
290
81. We wish to test H0: 21 µµ = versus Ha: 21 µµ ≠ Unpooled:
With Ho: 021 =− µµ vs. Ha: 021 ≠− µµ , we will reject Ho if α<− valuep .
( )( ) ( )
1695.15
1113
2
1252.12
1479.
2
1252.1
1479.
22
22
≈=
+
+=ν , and the test statistic
97.14869.
96.36.948.8
1252.1
1479. 22
−=−
=+
−=t leads to a pvalue of 2[ P(t > 1.97)]
( ) 062.031.2 ≈≈
Pooled:
The degrees of freedom 24212142 =−+=−== nmν and the pooled variance
is ( ) ( ) 3970.152.12411
79.2413 22 =
+
, so 181.1=ps . The test statistic is
1.2465.
96.
181.1
96.
121
141
−≈−
=+
−=t . The pvalue = 2[ P( t24 > 2.1 )] = 2( .023) = .046.
With the pooled method, there are more degrees of freedom, and the pvalue is smaller than with the unpooled method.
82. Because of the nature of the data, we will use a paired t test. We obtain the differences by subtracting intake value from expenditure value. We are testing the hypotheses H0: µd = 0 vs
Ha: µd ? 0. Test statistic 88.3757.1
7197.1
==t with df = n – 1 = 6 leads to a pvalue of 2[ P( t >
3.88 ) ˜ .004. Using either significance level .05 or .01, we would reject the null hypothesis and conclude that there is a difference between average intake and expenditure. However, at significance level .001, we would not reject.
83.
a. With n denoting the second sample size, the first is m = 3n. We then wish
( )nn
4003900
58.2220 += , which yields n = 47, m = 141.
b. We wish to find the n which minimizes ( )nn
z400
400900
2 2/ +−α , or equivalently, the
n which minimizes nn
400400
900+
−. Taking the derivative with respect to n and
equating to 0 yields ( ) 0400400900 22 =−− −− nn , whence ( )22 40049 nn −= , or
0000,64032005 2 =−+ nn . This yields n = 160, m = 400 – n = 240.
Chapter 9: Inferences Based on Two Samples
291
84. Let p1 = true survival rate at Cο11 ; p2 = true survival rate at Cο30 ; The hypotheses are
0: 210 =− ppH vs. 0: 21 ≠− ppH a . The test statistic is ( )nmqp
ppz
11
21
ˆˆ
ˆˆ
+
−= . With
802.9173ˆ1 ==p , and 927.
110102ˆ 2 ==p , 871.
201175ˆ ==p , 129.ˆ =q .
( )( )( )91.3
0320.125.
129.871.
927.802.
1101
911
−=−
=+
−=z . The pvalue =
( ) ( ) 0003.49.391.3 =−Φ<−Φ , so reject Ho at any reasonable level. The two survival rates appear to differ.
85.
a. We test 0: 210 =− µµH vs. 0: 21 ≠− µµaH . Assuming both populations have
normal distributions, the twosample t test is appropriate. The approximate degrees of
freedom ( )
( ) ( )4.11
110102083.
70325125.
042721.22
2
=+
=ν , so we use df = 11.
437.411,0005. =t , so we reject Ho if 437.4≥t or 437.4−≤t The test statistic is
3.3042721.
68.≈=t , which is not 437.4≥ , so we cannot reject Ho. At significance
level .001, the data does not indicate a difference in true average insulinbinding capacity due to the dosage level.
b. Pvalue = 2P( t > 3.3) = 2 (.004) = .008 which is > .001.
86. ( ) ( ) ( ) ( )[ ]
41111
ˆ4321
244
233
222
2112
−+++−+−+−+−
=nnnn
SnSnSnSnσ
( ) ( ) ( ) ( ) ( )[ ] 2
4321
244
233
222
2112
41111
ˆ σσσσσ
σ =−+++
−+−+−+−=
nnnnnnnn
E . The estimate for
the given data is ( ) ( ) ( ) ( )[ ]
409.50
1225.112601.76561.174096.15=
+++=
Chapter 9: Inferences Based on Two Samples
292
87. 00 =∆ , 1021 == σσ , d = 1, nn142.14200
==σ , so
−Φ=
142.14645.1
nβ ,
giving =β .9015, .8264, .0294, and .0000 for n = 25, 100, 2500, and 10,000 respectively. If
the si 'µ referred to true average IQ’s resulting from two different conditions, 121 =− µµ
would have little practical significance, yet very large sample sizes would yield statistical significance in this situation.
88. 0: 210 =− µµH is tested against 0: 21 ≠− µµaH using the twosample t test,
rejecting Ho at level .05 if either 131.215,025. =≥ tt or if 131.2−≤t . With 20.11=x ,
68.21 =s , 79.9=y , 21.32 =s , and m = n = 8, sp = 2.96, and t = .95, so Ho is not rejected. In the situation described, the effect of carpeting would be mixed up with any effects due to the different types of hospitals, so no separate assessment could be made. The experiment should have been designed so that a separate assessment could be obtained (e.g., a randomized block design).
89. 210 : ppH = will be rejected at level α in favor of 21: ppH a > if either
645.105. =≥ zz . With 10.ˆ2500250
1 ==p , 0668.ˆ2500167
2 ==p , and 0834.ˆ =p ,
2.40079.0332.
==z , so Ho is rejected . It appears that a response is more likely for a white
name than for a black name.
90. The computed value of Z is 34.14634
4634−=
+−
=z . A lower tailed test would be
appropriate, so the pvalue ( ) 05.0901.34.1 >=−Φ= , so we would not judge the drug to be effective.
Chapter 9: Inferences Based on Two Samples
293
91. a. Let 1µ and 2µ denote the true average weights for operations 1 and 2, respectively. The
relevant hypotheses are 0: 210 =− µµH vs. 0: 21 ≠− µµaH . The value of the
test statistic is ( )
( ) ( )43.6
318083.7
39.17
30672.3011363.4
39.17
3096.9
3097.10
63.141924.140222
−=−
=+
−=
+
−=t .
The d.f. ( )
( ) ( )5.57
2930672.3
29011363.4
318083.722
2
=+
=ν , so use df = 57. 000.257,025. ≈t ,
so we can reject Ho at level .05. The data indicates that there is a significant difference between the true mean weights of the packages for the two operations.
b. 1400: 10 =µH will be tested against 1400: 1 >µaH using a onesample t test
with test statistic m
s
xt
1
1400−= . With degrees of freedom = 29, we reject Ho if
699.129,05. => tt . The test statistic value is 1.100.224.2140024.1402
3097.10
==−
=t .
Because 1.1 < 1.699, Ho is not rejected. True average weight does not appear to exceed 1400.
92. ( )nm
YXVar 21 λλ+=− and X=1̂λ , Y=2λ̂ ,
nmYnXm
++
=λ̂ , giving
nm
YXZ
λλ ˆˆ +
−= . With 616.1=x and 557.2=y , z = 5.3 and pvalue =
( )( ) 0006.3.52 <−Φ , so we would certainly reject 210 : λλ =H in favor of
21: λλ ≠aH .
93. 62.11̂ == xλ , 56.2ˆ2 == yλ , 77.1
ˆˆ21 =+
nmλλ
, and the confidence interval is
( )( ) ( )59.,29.135.94.77.196.194. −−=±−=±−
Chapter 9: Inferences Based on Two Samples
294